The h Bar

Sir Cumference

00:15

physics needs to be kept pure

it's like how people complain about bad grammar becoming more standard

Slereah

06:09

morning

Slereah

07:25

Hm

How to prove geometric optics

Apparently the rigorous process involves the Luneberg-Kline asymptotic expansion

But it seems long and tedious

Slereah

07:46

"Every mathematician knows it is impossible to understand an elementary course in thermodynamics."

:D

Slereah

08:06

"According to Gibbs, the geometrical structure of thermodynamics is described by a contact manifold, equipped with the contact form, whose zeroes define the laws of thermodynamics $$dE = T dS - P DV$$

the horror

"Substances are Legendre submanifolds of the Gibbs manifold"

"Legendre singularities and the icosahedron"

Help

This article keeps getting worse

Slereah

08:54

ncatlab.org/nlab/show/…

Least useful article

Emilio Pisanty

09:06

@DanielSank he said, unassumingly?

Official congratulations to @DanielSank for his role in the (now-released) Quantum Supremacy paper by the Google team!

15

Mithrandir24601

09:35

@SirCumference ha! It's not 'bad grammar' becoming standard, it's people's understanding of linguistics being horribly outdated

@EmilioPisanty and hooray! Quantum Computers can officially do things! :D

skullpatrol

hip hip hooray!

Slereah

10:22

English grammar is nothing but bad grammar of previous incarnations of the language

bolbteppa

Geometric optics is the theory of electromagnetic waves which are such that their amplitude and direction of the solution of the wave equation remains constant over any local region, so that the solution reduces to a plane waves locally, the condition for this to hold is that the wavelength goes to zero so that the wave can always be considered plane over any local region.

The mathematical way to say this is that asolution $f = A e^{i\psi}$ of the wave function is such that $\psi = \psi_0 + \frac{\partial \psi}{\partial \mathbf{r}} \cdot d \mathbf{r} + \frac{\partial \psi}{\partial t} dt + ... = a_0 + \mathbf{k} \cdot d \mathbf{r} - \omega dt = a_0 - k_{\mu} dx^{\mu}$ i.e. the phase is a plane wave to first order, and this dominates, so that the first derivative $k_{\mu}$ of the phase $\psi$ satisfies $k_{\mu} k^{\mu} = 0$, which is the Eikonal equation.

You can also get this by plugging $f$ into the wave equation $\partial_{\mu} \partial^{\mu} f = 0$ and neglecting all terms except the $k_{\mu} k^{\mu}$ term which again gives the Eikonal equation.

bolbteppa

10:41

wave equation* not wave function gah

What is rigor going to add to this

Slereah

10:59

The hard part is showing that $E \approx e^{i\psi}$ is a decent approximation, though

Ultradark

why is $|\Psi|^2$ proportional to probabilities?

Slereah

@Ultradark That is how the theory was constructed

$|\Psi|^2$ obeys all the rules of a probability distribution

Therefore it can be interpreted as such

Ultradark

can you normalize a pdf with a non-constant factor?

Slereah

I mean if it's not constant it's not really a factor

Ultradark

can you normalize a pdf with a function? (I mean)

why is it always a "normalizing constant"

Slereah

11:08

Because the Hilbert space is a vector space

There's no concept of multiplying two wavefunctions together

Also if you have to normalize it with another function, why not use the result as your wavefunction to start with

what's the point of the first non-normalized wavefunction

Ultradark

i was just asking in general, but I guess that makes sense physically

Slereah

The point of the Hilbert space is that it's the minimal mathematical structure to get the results of QM

it's a vector space because of the superposition principle

It has an inner product to give you a probability distribution

Ultradark

do you know what's going on here?

It looks like the points at infinity are being "blown up" to get from the LHS to the RHS

skullpatrol

what does the text that accompanies the diagrams suggest?

Ultradark

I'm gathering that there's some conformal map that sends minkowski space to ADS

What does 3 dimensional Minkowski space conformally compactified to a cube look like?

skullpatrol

11:26

sounds like an exercise left to the reader :P

Ultradark

Do you know the answer?

skullpatrol

nope, sorry

Ultradark

There's no pictures of this, I'm kinda surprised

they are all 2d

Slereah

If you want conformal compactification to a cube just use some basic $(0,1)$ to $\mathbb{R}$ diffeomorphism

Like $\arctan(x)$

Ultradark

in a two 2d compactification there are 4 points at infinity

am I right to say there will be 8 points at infinity in a 3d compactification?

Slereah

11:44

Why 8?

Ultradark

because the cube has 8 corners

Slereah

Oh

bolbteppa

$f(t,\mathbf{r}) = A(t,\mathbf{r})e^{i\psi(t,\mathbf{r})}$ is just a complex number representation of the solution of the wave equation $\partial_{\mu} \partial^{\mu} f = 0$ so there's no approximation (for the potential $A_{\mu}$ not $E$)

Slereah

Isn't it the $n-1$ polytope rather than just the corners?

bolbteppa

The approximation comes from assuming $A$ and $\psi$ have certain properties

0

Q: Spin J $A_J(s,t) = - \frac{g^2(-s)^J}{t-M^2}$ Amplitude?

In GSW equation (1.1.2) they define the scattering amplitude for a spin $J$ particle at high energies as $$A_J(s,t) = - \frac{g^2(-s)^J}{t-M^2}$$ mentioning it is an asymptotic approximation to a formula involving Legendre polynomials, and in this stackexchange post it is also defined, and in Wil...

quantum-field-theory

No votes for the origin of string theory :(

Ultradark

11:54

@Slereah It very well could be, why do you think that?

Slereah

Dunno

vague intuition

Ultradark

I don't think it's that because let's say 3d Minkowski is situated inside a regular octahedron instead

Slereah

Well I mean, usually the higher dimensional conformal compactification aren't done as a polygon, really

Usually you compactify it in spherical coordinates

So that the "canonical" diagram is just the usual diagram $\times S^{n-2}$

Let's see if there's a standard compactification of $\mathbb{R}^2$ as a square

Ultradark

can you give me an example of such a canonical diagram ? (for $\times S^{n-2}$)

Slereah

Well the canonical one is the Minkowski metric in spherical coordinates

$$ds^2 = -dt^2 + dr^2 + r^2 d\Omega^2$$

Switch to null radial coordinates

$$ds^2 = -dudv + (\frac{v - u}{2})^2d\Omega^2$$

Then use the conformal factor $[(1+u^2)(1+v^2)]^{-1}$

$$ds^2 = -\frac{dudv}{(1+u^2)(1+v^2)} + \frac{(v - u)^2}{2(1+u^2)(1+v^2)}d\Omega^2$$

Do a coordinate change with $u = \tan(p)$, $v = \tan(q)$

and then $T = (p+q) / 2$ and $R = (q-p) / 2$

You end up with $$ds^2 = -dT^2 + dR^2 + \sin^{n-2}(R) d\Omega$$

Or something like that

It's actually more $\mathbb{R} \times S^3$ if you consider the maximally extended spacetime, but the conformal compactification of Minkowski itself is just $\mathbb{R}^2 \times S^2$

That's the fairly standard conformal compactification that you see on every diagram of Minkowski space

probably not unique, though, but I'm not sure there are others that are as palatable

Ultradark

12:12

I don't understand why maximally extended spacetime would be $\Bbb R\times S^3$

Slereah

@Ultradark just look at the metric

$dr^2 + \sin^{2}(r) d\Omega$ is the metric of a $3$-sphere

Ultradark

if that's the case then wouldn't it be impossible to visualise in 3d?

Slereah

Certainly, though 1) this is in 4 dimensions 2) this is why you suppress the angular degrees of freedom on diagrams

in 2+1 dimensions it would be $dr^2 + \sin(r) d\theta$ which is... also probably a sphere, but not the standard metric for one?

So you have $\mathbb{R} \times S^2$

Ultradark

perfect

Slereah

I'm not sure there are other decent compactification of Minkowski space that have properties as nice as the Penrose one, though

The huge benefit of such compactification is that the radial part has the same causal structure as Minkowski space

so it's extremely easy to get the causal structure at a glance

Ultradark

12:24

okay. you could map the version inside the sphere to the cube because they are homeomorphic

Slereah

Errrrr

I mean I suppose, yes?

Although that's the surface of a cube

And not a cube itself

This isn't $\mathbb{R}^3 \to (0,1)^3$

Ultradark

yeah, so starting with $S^2$ with the causal structure of maximally extended Minkowski spacetime (supressing angular degrees of freedom) in the bulk of $S^2$

Slereah

Also if you map the sphere to a cube you're gonna lose the radial ~ Minkowski structure

Ultradark

are you sure about that?

Slereah

No, but I would guess it to be likely

Ultradark

12:40

doesn't the 2d maximally extended compactified spacetime have the radial structure?

Slereah

Yes, but it's a very simple spacetime

the metric is just $$ds^2 = -dT^2 + dR^2$$

Fairly easy to see it's just Minkowski spacetime

Ultradark

@Slereah thanks for the help with this

Sir Cumference

13:29

Does anyone remember when math was simple

Before we had manifolds we had shapes

And the word "number" used to be meaningful

All right I gotta wake up

Slereah

@SirCumference where do you put the cut off line

Remember in Ancient Greece when they murdered a guy because he proved that the square root wasn't a rational number

Sir Cumference

Probably when something as basic as addition leads to groups and other things

Slereah

I mean really I think people underestimate how nasty old timey math used to be

Sir Cumference

@Slereah Er, no I actually don't

Slereah

Simpler math also meant much more complicated proofs for some theorems

Sir Cumference

13:32

That's pretty insane

Slereah

Hippasus

Hippasus of Metapontum (; Greek: Ἵππασος ὁ Μεταποντῖνος, Híppasos; c. 530 - c. 450 BC), was a Pythagorean philosopher. Little is known about his life or his beliefs, but he is sometimes credited with the discovery of the existence of irrational numbers. The discovery of irrational numbers is said to have been shocking to the Pythagoreans, and Hippasus is supposed to have drowned at sea, apparently as a punishment from the gods for divulging this. However, the few ancient sources which describe this story either do not mention Hippasus by name (e.g. Pappus) or alternatively tell that Hippasus drowned...

"It is related to Hippasus that he was a Pythagorean, and that, owing to his being the first to publish and describe the sphere from the twelve pentagons, he perished at sea for his impiety, but he received credit for the discovery, though really it all belonged to HIM (for in this way they refer to Pythagoras, and they do not call him by his name)."

Sir Cumference

A punishment from the gods for discovering irrational numbers

That's a sentence I didn't think I'd read

Slereah

The Pythagoreans were big on the universe being math

Irrationals didn't sit well with them

Ryan Unger

I understand constructible numbers

Those can be irrational

Slereah

How do you feel about algebraic numbers

Ryan Unger

13:44

I don’t like polynomials

Slereah

1 is a polynomial solution

bolbteppa

How about elliptic functions

Slereah

Those are not numbers

Sir Cumference

What is a number tho

I don't think it actually has a definition

Slereah

Well see above

each number set is constructed differently

and each of them can be constructed in a variety of ways

I think you have the ~number object~ in category theory if you want to define numbers independently of their constructions

IIRC natural number objects are basically just the Peano axioms

bolbteppa

13:58

0 := { }, 1 = { { } }, 2 = { { }, { { } } }, ... (I think)

Emilio Pisanty

@Mithrandir24601 depends on what you mean by "do things", right? quantum computers have been factoring numbers, in the lab, since at least 2001. Or did you mean useful things?

=P

Slereah

ncatlab.org/nlab/show/natural+numbers+object

@bolbteppa IIRC 0 is $\{ \varnothing \}$ in Russell's axioms

but it is $\varnothing$ in the von Neumann and the other construction

bolbteppa

Lets keep it within the realm of reality and not n-reality

Mithrandir24601

@EmilioPisanty I mean ' things that aren't trivial on a desktop pc' :P

bolbteppa

Right maybe it's 0 = {{ }}

Slereah

13:59

Well no, what you're describing is the von Neumann one, I think

The Russell axioms are much worse

$1$ is defined as the equivalence of every set of cardinality $1$

bolbteppa

@Slereah here steemit.com/mathematics/@sinbad989/…

Slereah

Although then again Russell has hierarchic sets so maybe it's not that bad

bolbteppa

0 = { }, 1 = {0} = { { } } , 2 = {0,1} = { {}, {{}} }, ...

Slereah

1 is like $$1 = \{ A | \forall x \in A, \not\exists y, x \neq y \wedge y \in A \}$$

yuvraj singh

How you define a mass in space, because there is no inertia in space, because of no force.

bolbteppa

14:02

So $1 \in 2$

and $0 \in 1$ (take that $0 = 1$ examples like $x = y \mapsto 0 = 1$!)

Slereah

Yeah that is the benefit of the von Neumann axioms

Emilio Pisanty

@bolbteppa $1!!\neq 0$

Slereah

since you have $a < b \sim a \in b$

Random-15

Hello, can anyone tell me why we can't use reduced mass in pulley system?

I let's say we have a pulley, on one end of the string we have one object of mass m_1, and the other end of mass m_2.

Then the tension in string comes out to be \frac{2m_1m_2}{m_1+m_2}, but lets say we take reduced mass, and consider the other object stationary.

Ryan Unger

@SirCumference of course they do

Random-15

14:06

In that case, we have tension equal to half of actual tension.

Ryan Unger

Christodoulou starts his physics classes with the definition of numbers because it’s so important

And physicists don’t know these things

Slereah

We learned about the numbers

Back in kindergarten

1

2

possibly others

bolbteppa

Using apples and slices of pizza

Slereah

Those are graded algebras @bolbteppa

Different matter

Random-15

This must be definitely because it's in respect to mass m_1. So when taking with respect to ground, what happens such that acceleration becomes double?

I tried to find out the answer but I couldn't understand.

I thought it should be half of what reduced mass gives, but it's actually double.

I asked my teacher but he said that he'll teach properly with center of mass, but I want to know the answer.

Slereah

14:16

ncatlab.org/nlab/show/sheafification

I refuse to believe that is a word

Ryan Unger

Why

Slereah

Very silly

Ryan Unger

It does exactly what you think it does

Slereah

Unlikely, since I don't know what a sheaf is

I don't know how to make something more sheaf-like than previously

Ryan Unger

Well that article is a bit incomprehensible

But check Wells’ book

Slereah

14:19

finding a good category theory ressource ain't easy

Aaron Stevens

@yuvrajsingh Why do you think forces do not exist in space?

Slereah

Though I'm thinking of getting that spivak book

mitpress.mit.edu/books/category-theory-sciences

Ryan Unger

Sheaves of rings/Abelian groups are not hard to understand

Slereah

I hear it's alright?

I vaguely recall that it is basically mapping objects to open sets

or vice versa

since that's how AQFT works

bolbteppa

No real results in any of this stuff, just more words to define things you already know

Ryan Unger

14:26

Implying physics contains real results...

bolbteppa

QED laughs at you

Ryan Unger

QED is old physics

Hard to argue with that

Sir Cumference

@RyanUnger i don't think i've ever heard the word "number" come up in any of my undergrad math courses

Aside from the usual phrases like "natural numbers"

bolbteppa

Numbers only arise as combinatorial or normalization factors

Ryan Unger

In principle I agree with @bolbteppa but I’d rather defend number theorists than agree with a physicist

This pains me

Sir Cumference

14:31

number theory is a meme

bolbteppa

Still need to read Riemann's paper properly

yuvraj singh

A body in space, is like floating, I does not have frame of reference.

John Rennie

@yuvrajsingh hi

Ryan Unger

@SirCumference what does this mean?

Sir Cumference

@RyanUnger your respect for set theorists equates to my respect for number theorists :P

Jul 3 at 15:51, by Ryan Unger

set theory is a meme

Ryan Unger

14:40

Wow what an awful take

Sir Cumference

14:55

Number theory doesn't really help anyone

And it's not a beautiful field either

It's probably the only math that I can't manage to get any interest in

Ryan Unger

@SirCumference this is such an awful take

Jesus

DanielSank

@EmilioPisanty Thanks!

Slereah

Isn't number theory used in cryptography

Ryan Unger

I finally found someone with worse math tastes than me

Huzzah

Sir Cumference

@Slereah That's like the only cited application

but even as a pure math field, it's just not beautiful

like really how often has anyone said "holy crap that makes so much sense" regarding a number theory theorem

Slereah

15:10

@SirCumference i mean it's more applications than GR

Sir Cumference

yeah but GR is cool

ok i'm explaining this badly but still

Ryan Unger

Do you even know what number theorists do

Sir Cumference

they study $\mathbb{Z}$ and its properties right?

JMac

physics.stackexchange.com/questions/509647/… Ever notice how the less clear the question is, the more aggressively the OP's will respond to people questioning them?

Sir Cumference

OK i actually need to finish my analysis hw

Slereah

15:23

@RyanUnger do they find new numbers

2

Emilio Pisanty

@PM2Ring where's the python room?

I have some questions but they're probably too basic for main SO

Ryan Unger

@Slereah yep.

Finding large numbers is a hard problem

If you want to do it correctly

Slereah

Any python question can be asked here :

biology.stackexchange.com/questions/tagged/herpetology

Sir Cumference

15:51

@RyanUnger i dunno, in terms of enjoyment I'd rank math fields as analysis > logic/set theory > algebra > things i don't like > number theory

like if there's some really cool aspect about studying primes and other aspects of the integers, i'm just missing it

Slereah

quite lucky that it's rarely used in physics

Imagine if all particle masses had to be prime numbers

Sir Cumference

@Slereah i think we can say "quite lucky that's it's rarely used in [any subject]" aside from cryptography :P

Slereah

Too bad cryptography is so important now!

Sir Cumference

16:06

to be fair the fact that people study set theory as a career means it's gotta have some very interesting aspect i haven't gotten to, so i ought to keep an open mind

Slereah

Like 100 years ago almost nobody had any reason to give a shit about cryptography

Unless you were a SPY or something

And now we use it all the time

And even then I think old timey cryptography used very little number theory

it was mostly just substitution and transposition

It was usually enough in the pre-computer era

Earliest prime-based cryptography is 1874 apparently

Sir Cumference

@SirCumference Meant to say number theory, tired

@Slereah it's weird to think we're living in a post-computer era when it only really began a few decades ago

like pre-computer era was 99.99999999999% of human history and now suddenly everything is different

i dunno i'm sleepy

Slereah

well I mean a lot of things changed in recent human history, to be fair

Also like 90% of human history was just hunting animals and making stylish loincloths

so obviously yes things have changed a lot

ACuriousMind

@Slereah we still make stylish loincloths, so that's some continuity there

Slereah

nah they don't make them like they used to

Sha Vuklia

16:36

does anyone know where dimensional analysis is explained a bit "rigorously"? I know the computations are not hard, but I don't really understand from which principles we work. How can we tell from the Hamiltonian (without resorting to classical physics) that energy has dimensions mass*lenghte^2*time^-2?

Say we have a potential $V_0\delta(x)$. Then our Hamiltonian looks like $p^2/2m+V_0\delta(x)$. So.. how do we decide which are going to be our "base quantities”? I mean, I guess since $p=-i\hbar\partial/\partial x$ (in one dimension) we know the dimensions of $p$ if we know those of $\hbar$ and $x$. We can set the dimension of $x$ to be length, and we can include mass as well (since $m$ is in the Hamiltonian anyways)

However, from $\Delta E\Delta t\geq\hbar/2$, it would seem that we should also set $\hbar$ to something, but that’s not the case, since we know that $E=1/2mv^2$, so $E$ is fixed, and therefore $\hbar$. But.. how can I derive that, I guess?

in this expression of the expectation value of the kinetic energy, we still have something in terms of $\hbar$

I just wish it was explained somewhere consicely and rigorously, because whenever I have to do dimensional analysis, I just resort to random classical formulas to get something that I can recognise, but I kind of want to have an "algorithm" or an explanation (from first principles/rules) from which I can derive things if I've forgotten how it works

Slereah

I mean there isn't a whole lot of math in dimensional analysis

Sha Vuklia

I know that

Slereah

The most math you can use for dimensional analysis is graded algebras

Sha Vuklia

the math isn't the problem

the problem (for me) is identifying which things we are going to set as fundamental quantities/dimensions I guess, based purely from the equation we're working with

Slereah

It's usually based on the context really?

The same equations can have different dimensions

Sha Vuklia

16:45

well, that's why I gave an example hamiltonian right

$p^2/(2m)+V_0\delta(x)$

Slereah

Well it all depends on how you define everything

Sha Vuklia

that's my question exactly:x

I don't know what the thought process behind that is

Slereah

Usually a momentum is $\sim M \cdot L/T$

Sha Vuklia

well....

could we work with the operator definition from QM?

I kind of don't want to resort to classical physics:x

Slereah

Well at some point of your theory, you have to relate the operators you're using with real life measurements

Sha Vuklia

16:47

the reason for that is that in QM, momentum is given in terms of $\hbar$

Slereah

And that's where you can relate the two

I'm sure you can have two operators which are basically identical as far as mathematical properties go

Sha Vuklia

and I kind of don't want to pick random formulas, like $E=\hbar f$ to know what to do with $\hbar$

Slereah

But are different as far as units go

What you need is some basic idea of what you're measuring

Sha Vuklia

well say we want to give(/measure) the ground state energy

then dimensional analysis can give an indication what it will be

but I don't really know what to do with $\hbar$

Slereah

ie if you have an observable $\hat{A}$, which gives rise to the expectation value $\langle \hat{A} \rangle$, then $\hat{A}$ will have the dimension of that measurement

Well, not quite

since the wavefunction has a dimension, too

That one isn't too hard to work out

Sha Vuklia

16:49

okay, so our observable here is $\hat H$, for the sake of a context

@Slereah uh:x

bolbteppa

@ShaVuklia why can't you resort to classical physics when there has to be a classical limit for things like energy

Slereah

$$\langle \Psi, \Psi \rangle = \int \Psi^*(x) \Psi(x) d^nx 1$$

Sha Vuklia

ah, right okay

Slereah

Therefore $$[\Psi^2] = [L^n]^{-1}$$

Sha Vuklia

@bolbteppa I don't want to use random classical formulas. I feel like it's easy enough to get it from commutation relations and the hamiltonian itself

@Slereah right that makes sense

but does it really matter for the ground state energy? we have $H\psi=E\psi$ anyways

Slereah

16:51

But I mean dimensions are an arbitrary label in the end

bolbteppa

Where does the Schrodinger equation even come from without considering a classical limit

Sha Vuklia

so the dimensions of $\psi$ kind of don't matter right

Slereah

It's what meaning we apply to them that is important

Well it does!

Since $[\Psi] = [L^{-3/2}]$

Sha Vuklia

uh:x

right okay

but how does that help us to determine the ground state energy?

bolbteppa

Well I guess Dirac has a way but ....

Sha Vuklia

16:52

@bolbteppa well I'm kind of taking that as a starting point

(I want to start somewhere, and not just do random things, because I'm afraid I might otherwise end up in a circular reasoning)

bolbteppa

$\hbar$ has units of $[E][T]$ and so $i \hbar \partial_t \psi = \hat{H} \psi$ shows $\hat{H}$ has units of energy

Slereah

As I said really

What you want is just to relate the operators with the measurements

Otherwise it's all arbitrary

Sha Vuklia

@Slereah but I don't understand how that process goes:(

Slereah

The measurement of position is $$\langle \hat{x} \rangle$$

bolbteppa

@ShaVuklia there's only one way you can trust you wont commit circular reasoning in my mind

Course of Theoretical Physics

The Course of Theoretical Physics is a ten-volume series of books covering theoretical physics that was initiated by Lev Landau and written in collaboration with his student Evgeny Lifshitz starting in the late 1930s. It is said that Landau composed much of the series in his head while in an NKVD prison in 1938-1939. However, almost all of the actual writing of the early volumes was done by Lifshitz, giving rise to the witticism, "not a word of Landau and not a thought of Lifshitz". The first eight volumes were finished in the 1950s, written in Russian and translated into English in the late 1950s...

Slereah

16:54

Therefore $x$ has the unit of length

Sha Vuklia

@bolbteppa I guess this is fair

Slereah

@ShaVuklia It's one of those rarely brough up thing in physics

The linking of the mathematical formalism to the actual experiments

The rules of correspondance, as they are called

bolbteppa

@RyanUnger ask your new cool friends what they think of those books

Slereah

@RyanUnger Ask your cool new friends to scan that thesis

bolbteppa

@Slereah alumni library privileges give you a way to get it?

Novalium Company

16:57

Evening.

Sha Vuklia

Slereah

Who knows, but I tried and I really can't get it

Short of stealing it

Sha Vuklia

I think the only thing that confuses me now, is what to do with $\hbar$

Novalium Company

What's the word for the thing people vote in? Like, the word poll is only for elections. What's the word for any topic?

Sha Vuklia

we can say choose time and length and mass as your fundamental dimensions, but what do we do with $\hbar$?

bolbteppa

16:58

From $\psi \approx e^{iS/\hbar}$ we see $\hbar$ has units of energy times time so that the argument of the exponential is dimensionless since the action has units of energy times time

Sha Vuklia

yea, I'm convinced between the relation of $\hbar$ and energy and time

bolbteppa

via $S = \int [T - V] dt $

Sha Vuklia

but I feel like we have one degree of freedom left

Novalium Company

:52262680 So if I vote whether Justin Bieber to wear his pants up or down, the form will be called an election, not a poll or something else?

DanielSank

twitter.com/IvankaTrump/status/1186987509609385988

Sha Vuklia

16:59

If $\hat H=i\hbar\partial/\partial t$, then we either have to choose energy or $\hbar$ as a fundamental quantity, no?

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